Question

Prove that :
$\sf{\dfrac{cosA-sinA+1}{cosA+sinA-1} = cosecA+cotA$

General Instructions :
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Use G00GLE for answering but answer should be correct .

Answer 1

Answer:

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Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: \dfrac{cosA - sinA + 1}{cosA + sinA - 1} = cosecA \: + \: cotA$

$\large\underline{\bf{Solution-}}$

Consider,

$\rm :\longmapsto\:\dfrac{cosA - sinA + 1}{cosA + sinA - 1}$

☆ Divide numerator and denominator by sinA, we get

$\rm \:= \: \:\dfrac{\dfrac{cosA}{sinA} - \dfrac{sinA}{sinA} + \dfrac{1}{sinA} }{ \: \: \: \: \: \dfrac{cosA}{sinA} + \dfrac{sinA}{sinA} - \dfrac{1}{sinA} \: \: \: \: \: }$

$\rm \:= \: \:\dfrac{cotA - 1 + cosecA}{cotA + 1 - cosecA}$

$\: \: \: \: \: \red{\bigg \{ \sf \: \because \: \dfrac{cosx}{sinx} = cotx\: and \: \dfrac{1}{sinx} = cosecx\bigg \}}$

$\rm \:= \: \:\dfrac{cotA+ cosecA - 1}{cotA + 1 - cosecA}$

$\rm \:= \: \:\dfrac{cotA+ cosecA - ( {cosec}^{2}A - {cot}^{2}A)}{cotA + 1 - cosecA}$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \sf \because \: {cosec}^{2}x - {cot}^{2}x = 1 \bigg \}}$

$\rm \:= \: \:\dfrac{(cotA + cosecA) - (cosecA + cotA)(cosecA - cotA)}{cotA + 1 - cosecA}$

$\rm \:= \: \:\dfrac{(cotA + cosecA) \: \cancel{(1 - cosecA + cotA)}}{ \cancel{1 - cosecA + cotA}}$

$\rm \:= \: \:cosecA + cotA$

${{\boxed{\bf{Hence, Proved}}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1